Problem: Find $\lim_{x\to -1}\dfrac{x^2-9}{x^2+1}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-4$ (Choice B) B $-5$ (Choice C) C $-9$ (Choice D) D The limit doesn't exist
Answer: Let's try to find the limit using direct substitution. $\begin{aligned} \lim_{x\to -1}\dfrac{x^2-9}{x^2+1}&=\dfrac{(-1)^2-9}{(-1)^2+1} \\\\ &=\dfrac{-8}{2} \\\\ &=-4 \end{aligned}$ We got a finite number. Since $\dfrac{x^2-9}{x^2+1}$ is continuous across its domain, we can determine that $\lim_{x\to -1}\dfrac{x^2-9}{x^2+1}$ is indeed equal to $-4$. In conclusion, $\lim_{x\to -1}\dfrac{x^2-9}{x^2+1}=-4$.